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Theorem ddif 4136
Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)
Assertion
Ref Expression
ddif (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 vex 3477 . . . . 5 𝑥 ∈ V
2 eldif 3958 . . . . 5 (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥𝐴))
31, 2mpbiran 706 . . . 4 (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥𝐴)
43con2bii 357 . . 3 (𝑥𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
51biantrur 530 . . 3 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
64, 5bitr2i 276 . 2 ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥𝐴)
76difeqri 4124 1 (V ∖ (V ∖ 𝐴)) = 𝐴
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wa 395   = wceq 1540  wcel 2105  Vcvv 3473  cdif 3945
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3475  df-dif 3951
This theorem is referenced by:  complss  4146  dfun3  4265  dfin3  4266  invdif  4268  ssindif0  4463  difdifdir  4491
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